Is $(0,1]$ a closed or open set? Is $A=(0,1]$ a closed or open set?
I think it's not an open set because it is not a subset of its interior points. Mainly, $1\in A$ but $1\not\in A^\circ$.
If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 \in A^c$ but $0\not\in (A^c)^\circ$
 A: It's important that you specify where you are considering the subset $A$. 
If $A \subset X$ with $ X = \mathbb{R}$, J.G. is absolutely right in the usual topology of $\mathbb{R}$. 
If $A \subset X$ with $ X = [0,1]$, $A^c = \{0\} $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.  
A: Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-\infty,\,0]\cup (1,\,\infty)$, which doesn't contain a neighbourhood of $0$.
A: A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $\mathbb{R}$ with the usual topology. The empty set $\emptyset$ is always both open and closed, no matter what the ambient space is.
